Smoothing requirements after rectification: Which rectifier configuration delivers the least-ripple output for a given load and capacitor, thereby requiring the least additional filtering?

Introduction / Context:
After converting AC to pulsating DC, designers usually add filters to reduce ripple. The ripple frequency of the rectified output determines how large a capacitor or inductor is needed. Full-wave and bridge rectifiers double the ripple frequency compared with half-wave rectifiers, reducing filter requirements.

Given Data / Assumptions:

• Same load resistance and same filter capacitor across compared rectifiers.
• Input is a sine wave at mains frequency f.
• Ideal diodes for conceptual comparison.

Concept / Approach:
A half-wave rectifier passes only one half-cycle, producing ripple at frequency f. Both the center-tapped full-wave rectifier (two diodes) and the bridge rectifier (four diodes) produce pulses on both half-cycles, so ripple occurs at 2f. For the same load and capacitance, higher ripple frequency means less discharge time per cycle and therefore less ripple amplitude. Hence, full-wave and bridge rectifiers require less filtering than a half-wave rectifier.

Step-by-Step Solution:

Verification / Alternative check:
Use the approximate capacitor-input filter relation: ΔV ≈ I_load / (f_ripple * C). Doubling f_ripple halves the ripple, demonstrating why full-wave/bridge are superior in this respect.

Why Other Options Are Wrong:

• Half-wave: Lowest ripple frequency, largest ripple, most filtering required.
• Full-wave alone or bridge alone: Each is correct individually; the best answer groups them because both exhibit 2f ripple.

Common Pitfalls:
Confusing diode count with output quality; more diodes in a bridge do not worsen ripple—voltage drop changes but ripple frequency still doubles.

Final Answer:
A full-wave rectifier and a bridge rectifier